In the related art, there are cases in which a sensor that detects a state of an object by changing a resistance value has a small change rate of the resistance value, such as one percent, corresponding to a state change. In the case of using such a sensor, a change of resistance value is converted to an electrical signal and then the electrical signal is amplified by an amplifier circuit. In the case where the change of the resistance value is converted to current, a current amplifier circuit is used. However, the current amplifier circuit in the related art, there is a problem in which most of bias current is consumed.
For example, it is known that a current mirror circuit is used as the current amplifier circuit in the related art. In the case of using this kind of current amplifier circuit, the change of the resistance value is converted to the current at first. Provided that the change rate of the resistance value is ΔR and the bias current is Ib, the converted current becomes: Ib (1+ΔR). In other words, the change of the resistance value is converted to the current in which signal current IbΔR is superimposed on the bias current Ib.
The converted current is received in the current mirror circuit and amplified by a multiple of a device size ratio. Here, when the device size ratio is K, the current amplified by the current mirror circuit is KIb(1+ΔR). In this case, a ratio between the bias current consumed in an entire circuit and signal current to be output (hereinafter referred to as “ratio between the bias current and the signal current”) is: KIbΔR/(Ib+KIb)=KΔR/(1+K). Since normally K is a value of one or larger, the ratio between the bias current and the signal current KΔR/(1+K) becomes ΔR or less. Thus, the signal current becomes K-fold, but the bias current is also increased, thereby not improving the ratio between the bias current and the signal current.
As a method of improving the above situation, there is a proposed method in which the ratio between the bias current and the signal current is improved by subtracting predetermined current Ib1 from the converted current Ib(1+ΔR) before being received in the current mirror circuit. According to such a current amplifier circuit, output of the current mirror circuit becomes: K (Ib−Ib1+IbΔR). Therefore, the ratio between the bias current and the signal current becomes: KIbΔR/{Ib+K (Ib−Ib1)}.
For example, in the case where Ib=200 μA, ΔR=0.01 (=1%), Ib1=197.8 μA, and K=80, total bias current becomes 376 μA (=200 μA+80×(200 μA−197.8 μA)), and the signal current becomes 160 μA (=80×200 μA×0.01). Therefore, the ratio between the bias current and the signal current is: 160/376=0.43.